Simplify the following expression and state the condition under which the simplification is valid. $n = \dfrac{7q^3 - 98q^2 + 315q}{q^3 + 4q^2 - 45q}$
Answer: First factor out the greatest common factors in the numerator and in the denominator. $ n = \dfrac {7q(q^2 - 14q + 45)} {q(q^2 + 4q - 45)} $ $ n = \dfrac{7q}{q} \cdot \dfrac{q^2 - 14q + 45}{q^2 + 4q - 45} $ Simplify: $ n = 7 \cdot \dfrac{q^2 - 14q + 45}{q^2 + 4q - 45}$ Since we are dividing by $q$ , we must remember that $q \neq 0$ Next factor the numerator and denominator. $ n = 7 \cdot \dfrac{(q - 5)(q - 9)}{(q - 5)(q + 9)}$ Assuming $q \neq 5$ , we can cancel the $q - 5$ $ n = 7 \cdot \dfrac{q - 9}{q + 9}$ Therefore: $ n = \dfrac{ 7(q - 9)}{ q + 9 }$, $q \neq 5$, $q \neq 0$